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2 years ago

Which terrestrial planet experiences the shortest year?

Physics

1 answer:

Masteriza [31]2 years ago

8 0

Being the planet closest to the sun, Mercury must be the one
with the shortest year. That's how gravity works.

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Calculate the total resistance in a series circuit made up of resistances of 3Ω, 4Ω, and 5Ω.

-BARSIC- [3]

The total resistance in a series circuit is equal to the sum of all resistors (R total = ΣRi).

R total = R1 + R2 + R3 = (3 + 4 + 5) Ω = 12 Ω

5 0

2 years ago

Sound waves cannot travel in

julsineya [31]

Answer:

The answer is D. A Vacumn

Explanation:

Sound cannot travel through a vacuum. A vacuum is an area without any air, like space. So sound cannot travel through space because there is no matter for the vibrations to work in.

5 0

2 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

2 years ago

If you get too many fouls, you may not be able to play in the rest of the game.

trapecia [35]

Answer:

In the NBA that's true

4 0

2 years ago

Read 2 more answers

An 87.6 g lead ball is dropped from rest from a height of 7.00 m. The collision between the ball and the ground is totally inela

bija089 [108]

Taking specific heat of lead as 0.128 J/gK = c

We have energy of ball at 7.00 meter height = mgh = $87.6*10^{-3}*9.81*7$

When leads gets heated by a temperature ΔT energy needed = mcΔT

= $87.6*10^{-3}*0.128*10^3$ ΔT

Comparing both the equations

$87.6*10^{-3}*9.81*7$ = $87.6*10^{-3}*0.128*10^3$ ΔT

ΔT = 0.536 K

Change in temperature same in degree and kelvin scale

So ΔT = 0.536 $^0C$

7 0

2 years ago